Giuseppe Liotta (University of Perugia)

Transcription

1 Drawing Graphs with Crossing Edges Giuseppe Liotta (University of Perugia)

2 Outline Motivation and research agenda - Graph Drawing and the planarization handicap - Mutzel s intuition and Huang s experiment A Turán-type problem on Right-Angle-Crossing graphs (RAC graphs) Algorithmic questions about RAC graphs Variants of RAC graphs Open problems

3 Motivation and research agenda

4 Graph Drawing algebraic description G = (V, E) V = {1, 2, 3, 4, 5, 6} in algorithm E = { (1,3) (1,6) (2,3) 1 (2,5) (2,4) (2,6) (3,5) (4,5) (4,6) } out 6 3 drawing 2 5 4

5 Graph Drawing algebraic description G = (V, E) V = {1, 2, 3, 4, 5, 6} in algorithm E = { (1,3) (1,6) (2,3) 1 (2,5) (2,4) (2,6) (3,5) (4,5) (4,6) } out 6 3 drawing the drawing must be readable

6 Readability and Crossings edge crossings significantly affect the readability (see, e.g., Sugiyama et al., Warshall, North et al., Batini et al., mid 80s) - confirmed by cognitive experimental studies (Purchase et al., ) rich body of graph drawing techniques assume the input is a planar/planarized graph

7 Example: thetsm framework V = {1, 2, 3, 4, 5, 6 } E = {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6) } Planarization Introduction of dummy vertices Compaction Crossing vertices removal

8 The planarization handicap for dense enough or constrained enough drawings, reducing the number of crossings is not enough to make them readable

9 The planarization handicap for dense enough or constrained enough drawings, reducing the number of crossings is not enough to make them readable since many edge crossings are often unavoidable, which ones are acceptable?

10 A pioneering intuition by Mutzel in a 2-layer drawing, small skewness is better than small number of edge crossings [SIAM J.Opt. 2001] 34 crossings: minimum skewness (number of edges whose deletion makes it planar) 24 crossings: minimum number of crossings

11 From intuition to empiricism How people read graphs Human experiments Theories on how people read graphs Optimization criteria Algorithms How people draw graphs

12 The experiment of Huang et al. Observations from eye tracking No crossings: eye movements were smooth and fast. Large crossing angle: eye movements were smooth, but a little slower. Small crossing angle: eye movements were very slow and no longer smooth (back-andforth movements at crossing points).

13 The experiment of Huang et al. observation confirmed by lab. experiments (p<0.05)

14 Crossings: a new point of view Optimization criteria Algorithms How people draw graphs α compute drawings of graphs where edges that form small angles are forbidden/minimised

15 Crossings: a new point of view Optimization criteria Algorithms How people draw graphs α compute drawings of graphs where edges that form small angles are forbidden/minimised

16 Crossing angle resolution: The research agenda study drawings with optimum crossing angle resolution (RAC graphs/drawings) study drawings whose crossing angle resolution is (at least) a given α, for 0< α < π/2 compute drawings that optimize the crossing angle resolution

17 Right Angle Crossing Graphs (RAC Graphs)

18 A Turán-type problem what is the maximum size of a RAC graph, i.e. a geometric graph such that any two crossing edges cross at right angles?

19 A Turán-type problem what is the maximum size of a RAC graph, i.e. a geometric graph such that any two crossing edges form π/2 angles? Theorem [Didimo, Eades, L., 2009 ] A RAC graph with n 4 vertices has at most 4n-10 edges. Also, for any k 3 there exists a RAC graph with n = 3k-5 vertices and 4n-10 edges.

20 A Turán-type problem what is the maximum size of a RAC graph, i.e. a geometric graph such that any two crossing edges form π/2 angles? Theorem [Didimo, Eades, L., 2009 ] A RAC graph with n 4 vertices has at most 4n-10 edges. Also, for any k 3 there exists a RAC graph with n = 3k-5 vertices and 4n-10 edges.

21 Crossing graph Let G be a geometric graph; the crossing graph G of G is the intersection graph of the (open) edges of G G * has A node for each edge of G An edge between two edges that cross

22 Lemma The bipartite property The crossing graph of a RAC graph is bipartite. e 1 e 3 e 6 e 4 e 5 e 2 Proof: Say e 1, e 2, e 3 e k are k edges such that: e 1 crosses e 2, Then: e 2 crosses e 3, e k-1 crosses e k, and e k crosses e 1. e 1 is parallel to e 3, e 5,, and e 2 is parallel to e 4, e 6,. Thus k is even and G * has no odd cycles. RAC graphs are quasi-planar. Thus m 6.5n-20

23 Edge coloring Using the BIPARTITE PROPERTY we can find 1. a red edge set E r 2. a blue edge set E b 3. a green edge set E g Such that 1. red edges correspond to isolated vertices of the crossing graph 2. blue edges cross only cross green edges 3. green edges only cross blue edges

24 The two planar subgraphs Red-Blue graph G RedBlue = (V, E Red U E Blue ) Red-Green graph G RedGreen = (V, E Red U E Green ) G RedBlue and G RedGreen are both planar. Thus m 6n-12

25 A closer look to the red edges 1. the edges of the external face are all red 2. if the RAC graph is maximal, each internal face of G RedBlue has at least two red edges

26 At least two red edges per internal face every internal face has at least three ears

27 At least two red edges per internal face if a face has only one red edge, one ear consists of blue edges u

28 At least two red edges per internal face at least one vertex is visible from the apex of the ear, contradicting the maximality of the RAC graph e e u f f it follows that every internal face of a maximal RAC graph has at least two red edges

29 Proof of the theorem (1) A straight-line RAC drawing with n 4 vertices has m 4n-10 edges Let G be RAC maximal Notation: ω = number of edges of the external boundary m r, m b, m g = number of red, blue, and green edges f rb = number of faces of the red-blue graph G redblue Assumption: m g m b

30 Proof of the theorem (2) each internal face of G redblue has at least 2 red edges and the external face of G redblue has ω red edges; also each edge is shared by at most 2 distinct faces 2m r 2(f rb 1) + ω m r f rb 1 + ω/2 By Euler s formula for planar graphs m r + m b n + f rb 2 m b n 1 ω/2

31 Proof of the theorem (3) m b n 1 ω/2 G redgreen has the same external face as G redblue and this face consists of ω edges m r + m g 3n 6 (ω 3) m r + m g 3n 3 ω m 4n 4 3/2 ω

32 Proof of the theorem (4) Two cases are possible: Case 1: ω 4 m 4n 10 Case 2: ω = 3 m 4n 4 3/2 ω consider the internal faces of G redblue that share at least one edge with the external face (fence faces) fence face there are at least 1 and at most 3 fence faces

33 Proof of the theorem (5) Two sub-cases are possible if ω = 3 : Sub-case 1: there is a fence face with at least 4 edges m r + m b 3n 6 1 m r + m b 3n 7 By assumption we have m g m b m r + m g 3n 7 and we had m b n 1 ω/2 ω = 3 m 4n 9.5 m 4n 10

34 Proof of the theorem (6) Sub-case 2: each fence face has 3 edges (in this case there are exactly three fence faces) α β α + β + γ 360 α < 90 γ β 90 and γ 90 at least 2 fence faces have 3 red edges (and the external face has 3 red edges) 2m r 2(f rb 3) + 3*3 m r f rb + 3/2

35 Proof of the theorem (7) m r f rb + 3/2 and we had m r + m b n + f rb 2 m b n 7/2 and we had m r + m g 3n 3 ω ω = 3 m 4n 9.5 m 4n 10

36 Proof of the theorem (8) for any k 3 there exists a straight-line RAC drawing with n = 3k-5 vertices and 4n-10 edges. Let G be the union of a maximal planar graph with k vertices and its dual (except the external face) The dual graph has 2k-5 vertices (black nodes) and hence the total number of vertices is n = 3k-5. The number of edges of G is m = (3k 6) + 3(2k 5) + (3k 6 3) = 12k 30 = 4n 10 the proof that G is a RAC graph relies on a result of Brightwell and E. R. Scheinerman (SIDMA, 1993)

37 Proof of the theorem (9) a primal-dual circle packing representation planar 3-connected graphs Vertices (primal and dual) are circles. Touching circles edge Dual edge crosses the primal at right angles.

38 Algorithmic questions about RAC graphs and their relatives

39 Recognition problem instance: graph G with at most 4n-10 edges question: can G be realized as a RAC graph?

40 Recognition problem instance: graph G with at most 4n-10 edges question: can G be realized as a RAC graph? Theorem [Eades-L., 2010] Deciding whether a graph can be realized as a RAC graph is NP-hard

41 Some natural research directions consider specific graph families and/or add constraints investigate variants of RAC graphs: 1. allow crossing angles smaller than π/2 2. allow bends along the edges 3. allow 1+2

42 Upward RAC digraphs Theorem [Angelini et al., 2009] Deciding whether an acyclic digraph is an upward RAC digraph is NP-hard G G* Each edge is replaced by a K4 gadget

43 Upward RAC digraphs Theorem [Angelini et al., 2009] Deciding whether an acyclic digraph is an upward RAC digraph is NP-hard 1. G is upward planar G* is straight-line upward RAC drawable 2. Upward planarity testing is NP-hard 3. The transformation of a digraph G into G* gives an immediate linear-time reduction from the upward planarity testing problem to the straight-line upward RAC drawability problem

44 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers.

45 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers. Theorem [Di Giacomo,Didimo, Eades, L., 2010] An ordering of the nodes can be realised as a BiRAC drawing if and only if has no fan crossings and no triple crossings.

46 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers. Theorem [Di Giacomo,Didimo, Eades, L., 2010] An ordering of the nodes can be realised as a BiRAC drawing if and only if has no fan crossings and no triple crossings. Fan crossing Triple crossing

47 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers. Theorem [Di Giacomo,Didimo, Eades, L., 2010] There is a linear time algorithm to test whether a bipartite graph is a BiRAC graph

48 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers. Theorem [Di Giacomo,Didimo, Eades, L., 2010] There is a O(n 2 logn)- time algorithm to compute a BiRAC drawing with the minimum number of crossings

49 BiRAC graphs A bipartite graph is a BiRAC graph if it has a straight-line RAC drawing in two layers. Theorem [Di Giacomo,Didimo, Eades, L., 2010] The problem of finding a maximum BiRAC subgraph is NP-hard

50 Variants of RAC graphs

51 RAC drawings with bent edges how many bends per edge to break the O(n) barrier?

52 RAC drawings with bent edges how many bends per edge to break the O(n) barrier? Theorem [Didimo,Eades,L.,2009] Every graph has a RAC drawing with three bends per edge

53 RAC drawings with bent edges technique 1 place all vertices on a line and use 45 slopes for middle edgesegments technique 2 compute an orthogonal drawing with at most one bend/edge and box nodes (Papakostas and Tollis, 2000) and then transform it

54 RAC drawings with bent edges Theorem [Didimo,Eades,L.,2009] A RAC drawing with n vertices and at most one bend per edge has O(n 4/3 ) edges Theorem [Arikushi, Fulek, Keszegh, Moric, Toth, 2010] A RAC drawing with n vertices and at most one bend per edge has at most 6.5n-13 edges

55 RAC drawings with bent edges Theorem [Didimo,Eades,L.,2009] A RAC drawing with n vertices and at most two bends per edge has O(n 7/4 ) edges Theorem [Arikushi, Fulek, Keszegh, Moric, Toth, 2010] A RAC drawing with n vertices and at most two bends per edge has less than 74.2n edges

56 Relaxing the crossing angle constraint can we beat the O(n) barrier by allowing non-orthogonal crossings?

57 Relaxing the crossing angle constraint can we beat the O(n) barrier by allowing non-orthogonal crossings? let α be a given angle such that 0<α< π/2 LAC-α drawings: the crossing angle is at least α EAC-α drawing: the crossing angle is exactly α the edges may/may not have bends

58 Relaxing the crossing angle constraint Theorem [Dujmovic, Gudmundsson, Morin, Wolle,2010] A straight-line LAC-α drawing with n vertices has at most (3n-6) π/α edges π α Theorem [Ackerman, Fulek, Toth,2010] A straight-line EAC-α drawing with n vertices has at most 3(3n-6) edges.

59 Relaxing the crossing angle constraint Theorem [Ackerman, Fulek, Toth,2010] An EAC-α drawing with n vertices and at most one bend per edge has at most 27n edges Theorem [Ackerman et al.,2010] An EAC-α drawing with n vertices and at most two bends per edge has at most 477n edges

60 Relaxing the crossing angle constraint Theorem [Di Giacomo, Didimo. L., Meijer, 2009] Every graph has a LAC-α drawing with one bend per edge v 0 v 1 v 2 v 3 v 4 v 5

61 Relaxing the crossing angle constraint Theorem [Di Giacomo, Didimo. L., Meijer, 2009] Every graph has a LAC-α drawing with one bend per edge v The red segments have slope at most 1/(c+1) c c v 1 c c v 2 c c v 3 c c v 4 c The blue segments have slope at least (c+1) The crossing resolution is at least α c v 5

62 Summary of Turán-type results RAC LAC-α EAC-α straight-line 1 bend/edge 2 bend/edge 3 bend/edge n( n 1) M = 4n 10 M 6.5n 13 M 74. 2n M = 2 π n( n 1) n( n 1) n( n 1) M ( 3n 6) M = M = M = α n( n 1) M 3(3n 6) M 27n M 477n M = 2 M denotes the maximum number of edges in the drawing

63 Summary of Turán-type results RAC LAC-α EAC-α straight-line 1 bend/edge 2 bend/edge 3 bend/edge n( n 1) M = 4n 10 M 6.5n 13 M 74. 2n M = 2 π n( n 1) n( n 1) n( n 1) M ( 3n 6) M = M = M = α n( n 1) M 3(3n 6) M 27n M 477n M = 2 M denotes the maximum number of edges in the drawing OPEN: establish tight bounds

64 Some algorithmic questions testing/drawing special families of RAC graphs (example: graphs of degree at most three, graphs having 4n-10 edges) testing/drawing graphs that admit a RAC/ LAC- α / EAC-α representation with at most one (two) bends per edge compute straight-line drawings where the smallest crossing i angle is maximized (or, for example, i 0 2 is minimized) k 1 = π ( α ) k